Question

Divide 600 in two parts such that 40% of one exceeds 60% of the other by 120.
A workman is paid * 150 for each day he works and is fined * 50 for each day he is absent
month of 30 days he earned 2100. For how many days did he remain absent?​

Answer 1

$\red{\textbf{Answer :- }}$

Step-by-step explanation:

1)$\textsf{Let the one part be 'x' and the other}$

$\textsf{part be 600 - x}$

$\textsf{Therefore, according to the question.}$

$\textsf{(x × 40)/100 = {(600 - x) × 60}/100}$

$\textsf{+ 120}$

$\textsf{On solving both L.H.S. and R.H.S. we get}$

$\textsf{2x/5 = {(600 - x)3}/5 + 120}$

$\textsf{2x/5 = (1800 - 3x)/5 + 120 }$

$\textsf{Taking L.C.M of (1800 - 3x)/5 + 120/1,}$

$\textsf{and solving it, we get}$

$\textsf{2x/5 = (1800 - 3x + 600)/5}$

$\textsf{2x + 3x = 1800 + 600}$

$\textsf{5x = 2400}$

$\textsf{x = 2400/5}$

$\textsf{x = 480 }$

$\LARGE\textsf{Cross check :-}$

$\textsf{1st part = 480 and 2nd part = 120}$

40%$\textsf{ of the first part exceeds }$60%

$\textsf{of the other by 120.}$

40 % of 480 = 60 % of 120 + 120

$\textsf{192 = 72 + 120}$

$\textsf{192 = 192}$

$\textsf{L.H.S. = R.H.S.}$

$\textsf{Hence proved.}$

$\LARGE\textsf{(2) Answer :-}$

$\textsf{Let the workman was absent for n days. }$

$\textsf{ Hence, he was worked for (30−n) days.}$

$\textsf{His total wage2100=150×(30−n)−50n }$

$\textsf{2100=4500−200n }$

$\textsf{ n=12}$

$\textsf{ Hence, he was absent for 12 days.}$